Find the overall heat transfer coefficient of the construction shown in Figure 6.2 if components 1 and 3 are brick 10 cm in thickness and the center one is stagnant air 5 cm in thickness. Additionally, the wall is plastered with 25 mm plaster on each side.

Solution

Using the data shown in Tables A5.4 and A5.5 in Appendix 5, we get the following list of resistance values:

1. Outside surface resistance = 0.044.

7. Inside surface resistance = 0.12.

Total resistance = 1.18 m2-K/W or U = 1/1.18 = 0.847 W/m2-K.

In some countries minimum U values for the various building components are specified by law to prohibit building poorly insulated buildings, which require a lot of energy for their heating and cooling needs.

Another situation usually encountered in buildings is the pitched roof shown in Figure. 6.3.

Using the electrical analogy, the combined thermal resistance is obtained from or which gives

total

ceiling roof

UrAc

UCAC UrAr

where | |

Ur |
= combined overall heat transfer coefficient for the pitched roof (W/m2-K). |

Uc |
= overall heat transfer coefficient for the ceiling per unit area of the ceiling |

(W/m2-K). | |

Ur |
= overall heat transfer coefficient for the roof per unit area of the roof |

(W/m2-K). | |

Ac |
= ceiling area (m2). |

Ar |
= roof area (m2). |

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